A projectile is launched from ground level with an initial velocity of 50 m/s at

Question

A projectile is launched from ground level with an initial velocity of 50 m/s at an angle of 37º above the horizontal (use g = 10 m/s2; choose positive direction up)

• What is the horizontal component of the initial velocity? (round to the nearest integer)

• What is the vertical component of the initial velocity? (round to the nearest integer)

• What is the maximum height reached by the projectile? (round to the nearest integer)

• How long does it take for this projectile to reach the maximum height? (round to one decimal)

• What are the components of the projectile’s velocity at maximum height? (round to the nearest integer) vx = and vy =

• How long does it take this projectile to complete its flight (i.e. how long between the launch and the time it falls on the level ground again)? (round to the nearest integer)

• What is the horizontal range of this projectile? (round to the nearest integer) .

• What is the magnitude of the projectile’s final velocity? (round to the nearest integer) .

• What angle does the final velocity vector make with the horizontal? (round to the nearest integer) degrees

Explanation / Answer

Given,

v0 = 50 m/s ; theta = 37 deg

1)We know that

vx = v0 cos(theta)

vx = 50 x cos37 = 39.93 m/s

Hence, vx = 39.93 m/s

2)vy = v0 sin(theta)

vy = 50 x sin37 = 30.09 m/s

Hence, vy = 30.09 m/s

3)we know that,

Hmax = vy^2/2g

Hmax = 30.09^2/2 x 9.81 = 46.15 m

Hence, Hmax = 46.15 m

4)T = sqrt (2H/g)

T = sqrt (2 x 46.15/9.81) = 3.07 s

Hence, T = 3.07 s

5)at max height, vertical velocity becomes zero and horizontal velocity remains same through out the motion

vx,vy = (39.93,0) m/s

6)T = 2t

T = 2 x 3.07 = 6.14 s

Hence, T = 6.14 s

7)R = vx T

R = 39.93 x 6.14 = 245.17 m

Hence, R = 245.17 m

8)vy = v0y + at

vy' = 9.81 x 3.06 = 30.01

v = sqrt (39.93^2 + 30.1^2) = 49.96 m/s

Hence, v = 49.96 m/s

9)theta = tan^-1(30.1/39.93) = 37 deg

Hence, theta = 37 deg

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